Problem: A bracket sequence is called regular if it is possible to obtain correct arithmetic expression by inserting characters + and 1 into this sequence. For example, sequences (())(), () and (()(())) are regular, while )(, (() and (()))( are not. Let's call a regular bracket sequence "RBS".

You are given a sequence s of n characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

You have to replace every character ? with either ) or ( (different characters ? can be replaced with different brackets). You cannot reorder the characters, remove them, insert other characters, and each ? must be replaced.

Determine if it is possible to obtain an RBS after these replacements.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

Each test case consists of one line containing s (2 ≤ |s| ≤ 100) — a sequence of characters (, ), and/or ?. There is exactly one character ( and exactly one character ) in this sequence.

Output

For each test case, print YES if it is possible to obtain a regular bracket sequence, or NO otherwise}. 

You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).

Example

Input


5
()
(?)
(??)
??()
)?(?


Output


YES
NO
YES
YES
NO

Note

In the first test case, the sequence is already an RBS.

In the third test case, you can obtain an RBS as follows: ()() or (()).

In the fourth test case, you can obtain an RBS as follows: ()().

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
#define int long long int
#define pb push_back
#define akash ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define miel min_element
#define mael max_element
int32_t main()
{		
	akash;
        int t;
        cin>>t;
        while(t--)
        {
                string s;
                int i,o=0,c=0;
                cin>>s;
                int n=s.length();
                if(n%2==1)
                {cout<<"NO\n"; continue;}
                if(s[0]==')' || s[n-1]=='(')
                {cout<<"NO\n"; continue;}
                for(i=0;i<n;i++)
                {
                        if(s[i]=='(')
                        o++;
                        else if(s[i]==')')
                        c++;
                }
                if(o>(n/2) || c>(n/2))
                {cout<<"NO\n"; continue;}
                cout<<"YES\n";
        }
}